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I am currently looking at a paper in which continuous differentiability of a closed plane curve $\gamma$ is implied by the following property: for any $\alpha>0$, there exists $r>0$ such that for all $x\in\mathbb{R}^2$, each component of $\gamma\cap B_r(x)$ is an $\alpha$-Lipschitz graph over some line.

I am familiar with Rademacher's theorem which states that local Lipschitz continuity implies differentiability almost everywhere, but have not come across the above implication and am not sure how to prove it. It seems similar to Rademacher's theorem except that we end up with continuous differentiability after requiring some sense of 'uniform' local Lipschitz continuity for any Lipschitz constant. Any pointers would be appreciated.

(My use of 'uniform' here refers to the same Lipschitz constant holding on every component of $\gamma\cap B_r(x)$, whereas in the usual definition of locally Lipschitz we allow different Lipschitz constants in different neighbourhoods.

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Nice problem. If a function $f$ is differentiable at one point $x_{0}$ with derivative $f^{\prime}(x_{0})=m$, then by rotating the coordinates axes you can assume that the graph of $f$ near the point $x_{0}$ is the graph of another function $g$ which is differentiable at the point $y_{0}$ with $g^{\prime}(y_{0})=0$, where $(y_{0},g(y_{0}))$ in the new coordinates is $(x_{0},f(x_{0}))$ in the old ones. For the same reason, if $f$ is continuously differentiable in a small interval near $x_{0}$, then given $\alpha>0$, $|f^{\prime}(x)-f^{\prime}(x_{0})|\leq\alpha$ for all $x$ near $x_{0}$ and so by rotating the coordinates axes you can assume that $|g^{\prime}(y)-0|\leq\alpha$ for all $y$ near $y_{0}$.

Now given your curve, consider a portion which is the graph of a Lipschitz function $f$. You need to prove that $f$ is differentiable everywhere. The idea is that if $$ \ell=\liminf_{x\rightarrow x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}<\limsup _{x\rightarrow x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}=L, $$ then no matter how you rotate axes if you take $\alpha